The displacement of a particle executing SHM is given y = 0.25 sin 200t cm. The maximum speed of the particle is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$y = 0.25 s i n 200 t$
$s p e e d = \frac{d y}{d t} = 0.25 \times 200 c o s 200 t$
$M a x s p e e d = 0.25 \times 200 = 50 c m s^{- 1}$

Suggest Corrections

Similar questions

y = 3 sin [6.28 (0.50 x - 50 t)]

(x \to in cm, y \to cm, t \to in second)

1 m s^{- 1}

1.57 m s^{- 2}

x

0.25 s i n 200 t

y = 0.25 sin 200 t c m

y = 0.30 sin (220 t + 0.64)

Join BYJU'S Learning Program

Select...